Question

A horizontal force of 5 N acts on a body of
mass 2 kg initially at rest. It starts moving on
the table having coefficient of friction = 0.2.
Calculate:
(i) Work done by the applied force in 5 s.
(ii) Work done by force of friction in 5 s.
(iii) Work done by the net force in 5 s.
(iv) Change in kinetic energy of the body in 5 s.
What do you conclude from this example?​

Answer 1

Explanation:

Please find attached answer

Hope you got it!

Answer 2

$\huge\sf{Given:}$

Mass m = 3 kg

Force F = 8 N

Kinetic friction coefficient μ = 0.2

Initial velocity , u = 0

$\huge\sf{Solution:}$

t = 10 s

According to the Newton’s law of motion:

$\tt \: a {}^{'} = \frac{f}{m } = \frac{8}{3} = \tt\bold{\boxed{2.6 m/s {}^{2} }}$

Friction force = μmg

$\tt\to 0.2 \times 3 \times -9.8$

$\tt \to-5.88 N$

Acceleration due to friction:

$\tt a {}^{"} = \frac{5.88}{3} = \bold{\boxed{- 1.96 m/s {}^{2} }}$

The total acceleration of the body $\tt a^{'} +a^{"}$

$\tt \to 2.6 + (-1.96)$

$\tt\to 0.64 m/s^2$

According to the equation of the motion ;

${\underline{\boxed{ \tt \: formula \: used \to \: s = ut + \frac{1}{2} at {}^{2}}}}$

$\tt \to0 + \frac{1}{2} \times 0.64 \times (10) ^{2}$

$\tt \to 32m$

(a) $\tt Wa = F \times s = 8 \times 32$

$\to\tt \<strong>boxed { 256 J}</strong>$

(b) $\tt Wt = F \times s$

$\tt = -5.88 x 32$

$\tt = -188 J$

• $\tt Net force = 8 + (-5.88)$

$\to\tt \<strong>boxed{ 2.12 N}</strong>$

(c) $\tt Wnet = 2.12 \times 32$

$\tt \<strong>boxed{ 68</strong> J}$

(d) ${\underline{\boxed{\sf \: Final \: velocity \to v = u + at }}}$

$\tt\to 0 + 0.64 \times10$

$\tt\to6.4 m/s$

${\underline{\boxed{ \sf \: change \: in \: kinetic \: energy = \frac{1}{2} mv {}^{2} }}}$

$\to \tt \frac{1}{2} \times 2(u {}^{2} - v {}^{2} )$

$\to \tt \: (6.4) {}^{2} - 0 {}^{2}$

$\tt \to\boxed{41 J}$